Home > Calculus > Limits at Infinity and Horizontal Asymptotes

What is the limit of #cosx# as x goes to infinity?

Answer 1

There is no limit.

Recall or Note:

#lim_(xrarroo)f(x) = L# if and only if

for every positived #epsilon#, there is an #M# that satisfies: for all #x > M#, #abs(f(x) - L) < epsilon#

As #x# increases without bound, #cosx# continues to attain every value between #-1# and #1#. So it cannot be getting and staying within #epsilon# of some one number, #L#,

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Cameron Alexander

Refer to explanation

Choose two sequences such as

#a_n=2n*pi# and #b_n=(2n+1)*pi# where

#lim_(n->oo) a_n=oo# and #lim_(n->oo) b_n=oo#

But #lim_(n->oo)cos(a_n)=cos(2pin)=1# and #lim_(n->oo)cos(b_n)=cos((2n+1)pi)=-1#

Hence there is no limit for #cosx# as #x->oo#

The theorem that was used for the proof is

**(Divergence Criterion for Functional Limits): Let #f:A→R# #f:A→R#, and let #c# be a limit point of #A#. If there exist two sequences #(x_n)# and #(y_n)# in #A# with #x_n≠c# and #y_n≠c#, and: #lim_(n->oo)x_n=lim_(n->oo) y_n=c# but #lim_(n->oo)f(x_n)≠lim_(n->oo)f(y_n)# then we can conclude that the functional #lim_(x→c)f(x)# does not exist.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

William Abdalla

The limit of cos(x) as x goes to infinity does not exist.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.